I have used the following method:

Step 1: Dividing the eq wid y²  to get  a(b-c)x²/y² + b(c-a)x/y + c(a-b) = 0

Step 2: Since for perfect square Discriminant = 0 , b² = 4 ac

            , and upon Solving, we get   ( ab+ bc - 2ac)² = 0

Step 3: This implies 2ac=b(a+c)  b= 2ac/a+c  a,b,c are in H.P